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A proton and and `alpha `- particle are accelerated through a potential difference of `100 V`. The ratio of the wavelength with the proton to that associated to that associated with an `alpha` - particle is |
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Answer» `lambda=h/sqrt(2mQV)` `rArr lambda prop 1/sqrt(mQ)` `rArr lambda_p/lambda_alpha=sqrt((m_alpha Q_alpha)/(m_p Q_p))` `=sqrt((4m_pxx2Q_p)/(m_pxxQ_p))` =`2sqrt2` |
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